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Physical Chemistry I for BiochemistsChem340Chem340

Lecture 37 (4/18/11)

Yoshitaka IshiiYoshitaka IshiiCh. 9.4‐8

Announcement

• HW10 due date is 4/25

• Exam 3 will be returned probably this Friday• Exam 3 will be returned probably this Friday

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Definition of ionic activities (Correction)

• a+ = (m+/m0)+ = (m+/m0)+where m+ is molality of  positive ion, m is the molality of  the compound dissolved, & m0 = 1mol kg‐1. 

• Same for a‐ =(m‐/m0)‐

a+ = (c+/c0)+,

where c+ is molarity of positive ion & c0 = 1mol /L

Alternatively,

Chemical potential of electrolyteµ = µ

0 + RT ln(a), where a = {a+

+ a‐‐ }

Now using molality of ions,a+ = x++ = (m+/m0)+ and a‐ =(m‐/m0)‐

h ( / ) ( / )• Thus, a =(m+/m0)

+ (m‐/m0)‐++ ‐ ‐

• We define  mean ionic molality m and mean ionic activity constant  as 

m (m+)

+(m‐)‐ = (m+) +(m‐)‐ &  ++ ‐ ‐

• µsolute=  µ =  µ0 + RT ln(a) = [ µ 0 + RT ln( + ‐)] + RT ln(m/m )= [ µ0 + RT ln(++‐ )] +  RT ln(m/m 0)

+ RT ln( )= µsolute

00 + RT ln(m/m 0) +  RT ln( )

Deviation from ideal ioncsolution

“Normal” standard state

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Coulomb Interactions• Coulomb Energy E

)(4

),( 1221

21 qqqq

qqE q1 q2

Q. How much is r in vacuum?

• Coulomb Potential

4 120

21 rr q1 q2r

r

qq

r 0

11 4)(

q1 1

r

• Energy (w’) required to fill a charge q1 on the surface of a sphere (radius R)

q1

R

q

R

qdqw

rr

q

0

2

1

00 84

1

R

(P215)

Gibbs Energy for Solvation (correction)

q2

Energy to accumulate a charge  of q in a sphere of a radium r

Permittivity for free space

r

qw

08

Energy for the same sphere but in a solvent

qw

2

8'

+q

)1( rrr08

1

1

8'

0

20

rSolvation r

qwwG

The difference due to solvation is

<  0

+q

‐

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Coulomb Interaction in Three Medium

r

q

4

+qIn vacuum

r04

r

q

OH 024

+q

In water

I i i l ti ith i t th I

H2O ~80

Q. For the same r, how much isthe interaction in H2O?

x 1/80

)exp(4

)exp(4

2/1

02

02

rAIr

q

rr

q

OH

OH

+q

‐

‐

+

+ ‐

+

In ionic aqueous solution with ion strength I

9.4 Calculating  from the Debey‐Hückel Theory – Ionic Screening and Ionic Strength (correction)

The potential of an isolated ion having a charge of ze in an dielectric medium at a location r from the charge is

On the other hand, the potential of an isolated ion having a charge of ze in an dielectric medium is

,

r

zer

rionisolated

04

)(

)exp(4

)( rze

rsolution

Ionic Strength,

where 

)p(4

)(0 rr

solution

solvent

rA kT

zzmLmNe

0

22322 2/)1000(2

Ionic Strength

molecules/kg solvent

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)exp()(/)()( rrrrfionisolatedsolution

Q.  So which is greater?  solution or isolated ion?

zzmLmNe

22

32 2/)1000(2

m=0

m=0.001

Q. Find rDH for 3 solutions

rDH = 1/ is called Debye‐Hückel screening length             f(1/ ) =e‐1 ~0.368

It is convenient to combine the 

solventr

A kTLmNe

0

)1000(2

m=0.01

m=0 1concentration  dependent terms as Ionicstrength:

i

iii

ii zmzmzzm

I 2222

2

1

2

m=0.1

Ionic Strength (Ex 9.2)

h h h f l/k f l

i

iii

ii zmzmzzm

I 2222

2

1

2

• What is the ion strength of 1 mol/kg of NaCl and Na2SO4 ?

•

• The principal ions of human blood plasma and

i

NaClI 22 11112

1 i

SONaI 2242 2112

2

1

• The principal ions of human blood plasma and their molal concentrations are

• Calculate the ionic strength of blood plasma. m

Na 0.14 m, m

Cl 0.10 m, m

HCO3 0.025 m.

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Debye‐Hückel Limiting Law

ezz

2

||)ln(

I1/2Derivation Raff (p416-428)

kTzz

r

08||)ln(

Q. Is > 1 or < 1 ?

Q. Why are the slopes different?

• µsolute= [ µ0 + RT ln(++‐‐)] +  RT ln(m/m 0)

+ RT ln( )= µsolute

00 + RT ln(m/m 0) +  RT ln( )

AgNO3

ZnBr2

Debey-Huckel “limiting” law is valid only for low ionic strength

Solvent is treated as a structure

CaCl2

-less dielectric medium Ions surrounded by structured water Assume point charge Ion has a volume Water coordinates to ion as complex Effective molality decreased

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• P9.14) Calculate the Debye–Hückel screening l th 1/ t 298 K i 0 00100 l ti flength 1/ at 298 K in a 0.00100 m solution of NaCl.

KatmkgL

molkgI solvent

r

298/

1091.2 11

110

Use this eq (9.31)to solve problems.(9.29) needs a factor m

The text is incorrect 108 1010

9.5 Chemical Eqilibrium in Electrolyte Solutions – Solubility ProductFrom Sec 8.13

• K =(aieq)i

• ai = i(ci/c0)  kT

ezz

2

8||)ln( (*)

We next consider dissociation of MgF2 in an aqueous solution: 

• MgF2(s) Mg2+(aq) + 2F‐(aq) 

Because the activity of the pure solid can be set to equal 1, 

Ksp =(aieq)i =aMg2+aF‐2 = (cMg2+/c

0) (cF‐/c0)2  3  =6.4x10‐9

kTr08

Q. Does the solubility depend on the ion strength? cMg2+ cF-

( )

sp i Mg2+ F Mg2+ F cF‐ = 2cMg2+.  Two unknown (cF‐ and )

● First assume  =1.  Solve cF‐ cF‐ =0.565 x10‐3 molL‐1

From c, ionic strength I & use (*)  =0.87.  

Solve cF‐ with  cF‐ = 0.68 x10‐3 molL‐1  (Repeat the process)

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• P9.21) Dichloroacetic acid has a dissociation constant of Ka = 3.32 x 10–2. Calculate the degree of dissociation for a 0.125 m solution of this acid (a) using the 0. 5 solutio of t is acid (a) usi g t eDebye–Hückel limiting law and (b) assuming that the mean ionic activity coefficient is one.

• HA H+ + A-

0.125 –m1 m1 m1

(a) Assume =1 Obtain m1.

Ksp =(aieq)i =aH+aA-

= (m1/m0) (m1/m0)2/{(0.125 – m1)/m0}= m1

2/(0.125 – m1) m1 =0.050 Then, revaluate ln new m1

Protein Salting in/Salting outP    >  P‐ + H+

Ksp =  (c2/c0

2) 2 = s2 2

• When ionic strength I is low,  < 1

Solubility of protein increases (Salting in)

• When ionic strength I is high,  > 1

Solubility of protein decreases (Salting out)

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Solubility of hemoglobin in the presence of (NH4)2SO4

Solubility high for a lower ionic strength due to (NH ) SOdue to (NH4)2SO4

Adding some salts stabilizesprotein in a solution

Solubility lower for a higher ionic strength

9.6 The Electrochemical Potential• Assume that a Zn electrod is partially 

immersed in an aqueous solution of ZnSO4. 

• Zn(s)  Zn2+(aq) + 2e‐

‐Whereas Zn2+ goes into solution, e‐ stays on the electrode.  

‐ Only small Zn dissolves into ions (~10‐14 

mol), producing ~ 1V potential between Zn and electrolyte. 

To transfer a charge dQ from the potential To transfer a charge dQ from the potential 1 to 2, the required work is 

dG = dw = (2 ‐ 1)dQ ,where dQ = zFdn and z is a charge number of an ion (z = 0, 1, 2 ..) and F is a charge of 1 mol of an electron in absolute value .    

(Faraday constant)

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Electrochemical Potential (continued)

• dG = dw = (2‐ 1)dQ ,

where dQ = zFdn.

dG ( )dQ zF( ) dn (9 44) dG = (2‐ 1)dQ = zF(2‐ 1) dn (9.44)

We define now electrochemical potential as

(9.46 text incorrect)

So the difference             is given as

zF~ dndG 12~~

F~~ 12

~~

F~~

In electrochemical reaction, the equilibrium is reached when 

Fz 1212 Fz 1212

0~ ii

ireactionG 0 ii

ireactionG

We choose (sol) =0 for solution. So

9.7 Electrochemical Cells and Half‐Cells zF~

CellHalf Cell

)solutioninion(~

For electron in a metal electrode,(z = ‐1ele=0)

Now, we consider an equilibrium,Mz+ + ze‐ M,

Identicalelectrolyte

)solutioninion(ii

FFeleele ~

where M denotes metal.  For this,

Zn  Zn2+ + 2e‐ Cu2+ + 2e‐ Cu

eleZMM z ~~~

zFzFF ZMM 00

zFz ZMeleZMM ~~

Correct text

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Equilibrium at a half cell

(9.53)

• For a metal (made of a pure element)at 1 bar in a standard eleZMM z ~~

state, its chemical potential should be 0. (In text MZ+

0 = 0  Not exact)

• Using (9.53) and 

)298,1(0 0 KbarZMM

zFz ZMeleZMM ~~~

zF ~

Fele ~

In a standard condition at 298K and 1 bar 

Cf.                           and  

zFMZM

zFzFMZM 0~

0~ 0 MM Fele ~

Standard Hydrogen Electrode

H+(aq) + e‐ 1/2H2(g)

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Calculating Potential for a Hydrogen Electrode

H+(aq) + e‐ 1/2H2(g)

22/1~)(~HeleH aq 0

0 /ln)( ffRTpgas

(Reference electrode)

Potential of the hydrogen electrode

Using a standard state ( =0 in electrolyte, 298K, 1 bar),

)ln(2/12/1)ln( 20

2/0

HHHHHH fRTFaRT

0/ln)( ffRTp gasgas

)ln(

2/1 2/12

02

0

2/

H

HHH

HH a

f

F

RT

F

For unit activities of all species aH+ = f = 1, the cell (electrode) has its standard potential:

0

2/1 002

0

2/0

FF

HHHHH

Convention for H+

0H+ = 0

9.8 Redox Reaction in Electrochemical Cells and the Nernst Equation

• Left(anode):

• Right (cathode) :

Zn(s)  Zn2+ (aq) + 2e‐

Cu2+ (aq) + 2e‐ Cu(s) Daniel Cell

• Overall:

• Anode: Red1Ox1+1e‐

• Cathode: Ox2 +2e‐Red2• Overall:

Zn(s) + Cu2+ (aq)  Zn2+(aq) + Cu(s)

Zn  Zn2+ + 2e‐ Cu2+ + 2e‐ Cu

• Overall:

2Red1 + 1Ox2 2Ox1 +1Red2Q. How do you get Greaction

using chemical potential?

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Nernst Equation for a Cell (Derivation)

(9.69)• For a reaction, n moles (n=2) of electron are transferred

)/ln(~~

~~~~

2220

20

22

CuZnCuZn

CuZnCuZnreaction

aaRT

G

Zn(s) + Cu2+ (aq)  Zn2+(aq) + Cu(s)

For a reaction, n moles (n 2) of electron are transferred from the cathode to anode.  The potential difference is  =cathode‐anode.  

(z= ‐1 for electron) (9.70)

By combining (9.69) & (9.79), we obtain

nFnzFG reaction )(

nFEa

RTa

RTG ZnCuZn

Znti

)ln(~~)ln( 2

20

2020

where we defined  electromotive force (emf) as E = .  When aZn2+ = aCu2+ = 1, E

0 is defined as ‐nFE0 = G0re action.

E = E0 –(RT/nF)ln(azn2+/aCu2+) (9.72)

(9.73)

nFEa

RTa

RTGCu

CuZn

Cureaction

)ln()ln(2

22

2

E = E0 – (RT/nF)lnQ

Nernst’s Equation for Whole and Half Cells

For a whole cell,  2Red1 2Ox1 +ne‐ ; 1Ox2 +ne‐ 1Red2

• E = E0 (RT/nF)lnQ [ ])(1

2Re2

1

dOX aaQ

Balance the charges

• E =  E0 – (RT/nF)lnQ [                           ]

At 298.15K, we obtain the equation called Nernst’s equation

E = E0 – (RT/nF)lnQ = E0 – (0.05916/n)logQ (9.74)

For a half cell reaction:  Oxn+ + ne‐ Red~~~0 G F~

)(1

22

1ReOxd aa

Q

• Eox/Red =(0OXn+  0

Red)/nF - (RT/nF)ln(aRed/aOXn+) (9.77)

= E0Ox/Re– (RT/nF)ln(aRed/aOxn+)

0)/ln(~~0

Re/ReRe00

Re

dOxdOxdOx

deleOxreaction

nFaaRT

nG

Fele

Activity of electrons is not involved